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Subspaces$W$ and $Z$ of $\mathbb R^4$ are generated by $\{(1,1,0,-1),(1,2,3,0),(2,3,3,-1)\}$ and $\{(1,1,0,-1),(1,2,3,4),(0,1,3,5)\}$, respesctively. Find a basis for $W$$\cap$$Z$.

I already know how to find the basis for $W+Z$, but I am confused on how to find the basis of $W$$\cap$$Z$.

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  • $\begingroup$ What is the word “matrix” doing in the title? $\endgroup$ – José Carlos Santos Oct 14 '18 at 13:47
  • $\begingroup$ Refer (math.stackexchange.com/questions/25371/…) $\endgroup$ – Chinnapparaj R Oct 14 '18 at 14:12
  • $\begingroup$ It might be expeditious to work out $\dim(W\cap Z)$ from $\dim(W),\dim(Z),$ and $\dim(W+Z)$, if you already have a basis for the latter. Note that the "generated by" vectors (spanning sets) for $W,Z$ may turn out not to be resp. bases for them (linearly independent as well as spanning)! $\endgroup$ – hardmath Oct 14 '18 at 15:39
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Here are some steps you can take to solve this problem. First, find bases of the orthogonal complements $W^\perp$ and $Z^\perp$ (of course, with respect to the usual nondegenerate bilinear form of $\mathbb{R}^4$). Then, we have $$(W\cap Z)^\perp = W^\perp + Z^\perp.$$ That is, $$W\cap Z=(W^\perp+Z^\perp)^\perp.$$

Now, to find a basis of $W^\perp$, write down the matrix $$w=\begin{pmatrix}1&1&0&-1\\1&2&3&0\\2&3&3&-1\end{pmatrix}$$ whose rows are the given vectors of $W$ that span $W$. Solve for the (right) null space of $w$. Similarly, write down the matrix $$z=\begin{pmatrix}1&1&0&-1\\1&2&3&4\\0&1&3&5\end{pmatrix}$$ whose rows are the given vectors of $Z$ that span $Z$. Solve for the null space of $z$.

If you did the job properly, you should see that $\operatorname{null}(w)$ is spanned by $(3,-3,1,0)$ and $(2,-1,0,1)$, while $\operatorname{null}(z)$ is spanned by $(3,-3,1,0)$ and $(6,-5,0,1)$. That is, $V=W^\perp +Z^\perp$ is spanned by $(3,-3,1,0)$, $(2,-1,0,1)$, $(3,-3,1,0)$, and $(6,-5,0,1)$ (the repetition of $(3,-3,1,0)$ can be removed). We now try to find $W\cap Z=V^\perp$.

Define the matrix $v$ by stacking up the known spanning elements of $V$: $$v=\begin{pmatrix}3&-3&1&0\\2&-1&0&1\\3&-3&1&0\\6&-5&0&1\end{pmatrix}.$$ Determine the null space of $v$, and the work is now yours.

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